Question

I was asked to write a swap without using temp variables or using xor and i came up with this.
In Java, this works, but in C/C++ this does not work.

I was under the impression that this would always work since the value of ‘a’ on the left side of the ‘|’ would be stored in a register and then the assignment to ‘a’ would occur negating the effect on the assigned value for ‘b’.

int a = 5;
int b = -13;
b = a | (0 & (a = b));

Answer 1

You are modifying a variable and reading its value without an intervening sequence point.

b =         a          + 0 *   (a = b);
//  reading a's value        modifying a

This is undefined behavior. You have no right to any expectations on what the code will do.